# Page:A Dynamical Theory of the Electromagnetic Field.pdf/9

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PROFESSOR CLERK MAXWELL ON THE ELECTROMAGNETIC FIELD.

momentum, which may be called the momentum of the fly-wheel reduced to the driving-point. The unbalanced force acting on the driving-point increases this momentum, and is measured by the rate of its increase.

In the case of electric currents, the resistance to sudden increase or diminution of strength produces effects exactly like those of momentum, but the amount of this momentum depends on the shape of the conductor and the relative position of its different parts.

Mutual Action of two Currents

(23) If there are two electric currents in the field, the magnetic force at any point is that compounded of the forces due to each current separately, and since the two currents are in connexion with every point of the field, they will be in connexion with each other, so that any increases or diminution of the one will produce a force acting with or contrary to the other.

Dynamical Illustration of Reduced Momentum

(24) As a dynamical illustration, let us suppose a body ${\displaystyle C}$ so connected with two independent driving-points ${\displaystyle A}$ and ${\displaystyle B}$ that its velocity is ${\displaystyle p}$ times that of ${\displaystyle A}$ together with ${\displaystyle q}$ times that of ${\displaystyle B}$. Let ${\displaystyle u}$ be the velocity of ${\displaystyle A}$, ${\displaystyle v}$ that of ${\displaystyle B}$, and ${\displaystyle w}$ that of ${\displaystyle C}$, and let ${\displaystyle dx}$, ${\displaystyle dy}$, ${\displaystyle dz}$ be their simultaneous displacements, then by the general equation of dynamics[1],

${\displaystyle C{\frac {dw}{dt}}\delta z=X\delta x+Y\delta y,}$

where ${\displaystyle X}$ and ${\displaystyle Y}$ are the forces acting at ${\displaystyle A}$ and ${\displaystyle B}$.

But

${\displaystyle {\frac {dw}{dt}}=p{\frac {du}{dt}}+q{\frac {dv}{dt}},}$

and

${\displaystyle \delta z=p\delta x+q\delta y.\,}$

Substituting, and remembering that ${\displaystyle dx}$ and ${\displaystyle dy}$ are independent,

 ${\displaystyle \left.{\begin{array}{l}X={\frac {d}{dt}}\left({Cp^{2}u+Cpqv}\right)\\Y={\frac {d}{dt}}\left({Cpqu+Cq^{2}v}\right)\\\end{array}}\right\}}$ (1)

We may call ${\displaystyle Cp^{2}u+Cpqv}$ the momentum of ${\displaystyle C}$ referred to ${\displaystyle A}$, and ${\displaystyle Cpqu+Cq^{2}v}$ its momentum referred to ${\displaystyle B}$; then we may say that the effect of the force ${\displaystyle X}$ is to increase the momentum of ${\displaystyle C}$ referred to ${\displaystyle A}$, and that of ${\displaystyle Y}$ to increase its momentum referred to ${\displaystyle B}$.

If there are many bodies connected with ${\displaystyle A}$ and ${\displaystyle B}$ in a similar way but with different values of ${\displaystyle p}$ and ${\displaystyle q}$, we may treat the question in the same way by assuming

${\displaystyle L=\sum {\left({Cp^{2}}\right)}}$, ${\displaystyle M=\sum {\left({Cpq}\right)}}$, ${\displaystyle N=\sum {\left({Cq^{2}}\right)}}$,
1. Lagrange, Mec. Anal. II, 2, 5.